4. (10 points) woman. If both disease traits are X-linked recessive what is the probability A man hemizygous for both hemophilia A and color blindness mates with a normal hemophilia A nor colorblindness if the two disease genes show complete that a mating between their children will produce a grandson with neither a. linkage? (5 points) that a mating between their children will produce a grandson with both hemophilia A and colorblindness if the two disease genes map 40 cM apart? (5 points)

4. (10 points) woman. If both disease traits are X-linked recessive what is the probability A man hemizygous for both hemophilia A and color blindness mates with a normal hemophilia A nor colorblindness if the two disease genes show complete that a mating between their children will produce a grandson with neither a. linkage? (5 points) that a mating between their children will produce a grandson with both hemophilia A and colorblindness if the two disease genes map 40 cM apart? (5 points)

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 7TYK: In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene...

See similar textbooks

Similar questions

A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
The genotype XXY corresponds to Klinefelter syndrome Turner syndrome Triplo-X Jacob syndrome
Human females have two X chromosomes XX; males have one X and one Y chromosome XY. a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. A female homozygous for an X-linked allele can produce how many types of gametes with respect to that allele? c. A female heterozygous for an X-linked allele can produce how many types of gametes with respect to that allele?
Analysis of X-Linked Dominant and Recessive Traits In the eighteenth century, a young boy with a skin condition known as ichthyosis hystrix gravior was identified. The phenotype of this disorder includes thickening of skin and the formation of loose spines that are sloughed off periodically. This man married and had six sons, all of whom had the same condition. He also had several daughters, all of whom were unaffected. In all succeeding generations, the condition was passed on from father to son. What can you theorize about the location of the gene that causes ichthyosis hystrix gravior?
The genotype XXY corresponds to: a. Klinefelter syndrome b. Turner syndrome c. Tripto-X d. Jacob syndrome
Assuming no gene linkage, in a dihybrid cross of AABB x aabb with AaBb F1 heterozygotes, what is the ratio of the F1 gametes (AB, aB, Ab, ab) that will give rise to the F2 offspring? a. 1:1:1:1 b. 1:3:3:1 c. 1:2:2:1 d. 4:3:3:1
A true-breeding rabbit with agouti (mottled, grayish brown) fur crossed with a true-breeding rabbit with chinchilla (silver) fur produces all agouti offspring. A true-breeding chinchilla rabbit crossed with a true-breeding Himalayan rabbit (white fur with pigmented nose, ears, tail, and legs) produces all chinchilla offspring. A true-breeding Himalayan rabbit crossed with a true-breeding albino rabbit produces all Himalayan offspring. Explain the inheritance of the fur colors.
A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of a color-blind father. Her mother had normal color vision, but her maternal grandfather was color-blind. What is the probability that your friend is color-blind? (a) 1 (b) (c) (d) (e) 0